How many grams ofPbCl2will dissolve in 2.00 L of O.650M aqueous Pb(N03)2 solution?

Answer 1

........PbCl2 ==> Pb^2+ + 2Cl^-
I.......solid......0........0
C.......solid......x........2x
E.......solid......x........2x

.........Pb(NO3)2 --> Pb^2+ + 2NO3^-
I........0.650M........0.......0
C.......-0.650.......0.650..0.650
E........0...........0.650..0.650

Ksp = (Pb^2+)(Cl^-)^2
Substitute Ksp value.
(Pb^2+) = x from PbCl2 + 0.650 from Pb(NO3)2
(Cl^-) = x
Solve for x.

Answer 2

The question asked for grams PbCl2.
x is in M, then mols = M x L and mols x molar mass = grams

Answer 3

@DrBob222 I understood up to the substitution part, can you explain that please?